Question

Determine the time at which the ball reaches its maximum height and find the maximum height

Answer 1

we have the function

`s(t)=-16t^2+20t+2`

this function represents a vertical parabola open downward

the vertex represents a maximum

Convert the given function to vertex form

so

y=a(x-h)^2+k

where

(h,k) is the vertex

step 1

Factor of -16

`s(t)=-16(t^2-(20t)/(16))+2`

step 2

Complete the square

`s(t)=-16(t^2-(20t)/(16)+(20^2)/(32^2)-(20^2)/(32^2))+2`
`s(t)=-16(t^2-(20t)/(16)+(20^2)/(32^2))+2+(16\cdot20^2)/(32^2)`

simplify

`s(t)=-16(t^2-(20t)/(16)+(20^2)/(32^2))+2+(25)/(4)`
`s(t)=-16(t^2-(20t)/(16)+(20^2)/(32^2))+(33)/(4)`

Rewrite as perfect squares

`s(t)=-16(t^{}-(20)/(32))+(33)/(4)`

simplify

`s(t)=-16(t^{}-(5)/(8))+(33)/(4)`

the vertex is the point (5/8,33/4)

therefore

the maximum height is the y-coordinate of the vertex

maximum height is 33/4=8.25 ft